Question 1

Why do restrictive enzymes that do blunt cuts even exist if they are so inefficient?

Accepted Answer

Restriction enzymes are found in bacteria and they have some biological role (explained below), but we are exploiting it in our way to use in experiment.

Biological role of restriction enzymes in bacteria: when restriction enzyme is present in a given bacterium, such bacterium can prevent (restrict) the growth of certain bacterial viruses (bacteriophages) and this is the reason also to call it as RESTRICTION enzymes. in this way it is the defensive enzyme that protects the host bacterial DNA from the DNA genome of foreign organism (bacteriophage) by specifically inactivating the invading bacteriophage DNA by digestion

Now one more question arises that WHY (& HOW) THESE RESTRICTION ENZYMES CAN CUT ONLY FOREIGN DNA BUT NOT THE HOST BACTERIUM'S DNA?
Explanation: if particular bacterium has restriction enzyme, it must have companion site specific DNA methylase which methylates DNA of host bacterium in site specific manner and methylated DNA is not the substrate for restriction enzyme. so host bacterium DNA is not cut by restriction but when new DNA is inserted by bacteriophage, it is not methylated and so it chopped by restriction enzyme and bacteria can survive (i.e. bacteria's innate immunity !)
Hope everything is clear...

Question 2

How long does the process of cutting DNA take?

Accepted Answer

It depends on the enzyme and the lab that produces them, but the rule of thumb for digestions is 1 hour at the appropriate temperature: For example, SmaI works at 25C, while EcoRI works at 37C.

If you want to know more then know that enzymes are sold in certain 'sizes', which are the units present on the vial. A unit, according to NEB is:

"One unit is defined as the amount of enzyme required to digest 1 µg of λ DNA (HindIII digest) in 1 hour at 25°C in a total reaction volume of 50 µl."

Hope this helps

Question 3

what would happen if the gap never closes?

Accepted Answer

Although the other answer is funnier, what would actually happen if the gap never closed during a ligation is that the DNA fragments would come apart again. The sticky ends will only hold them together briefly, and if ligase doesn't connect them during that time, they will go back to floating around and bumping into other pieces of DNA and enzymes in the reaction mix.

Question 4

I did not understand how to differentiate between plasmids in which the gene of interest has been correctly inserted and those in which it isn't.

Accepted Answer

You can read this article for more info: https://www.khanacademy.org/science/biology/biotech-dna-technology/dna-cloning-tutorial/a/bacterial-transformation-selection. I hope this helps you! :)

Question 5

How do scientists make sure that the bases of the plasmid are complementary to the bases of the inserted DNA?

Accepted Answer

The easy way is to use the same restriction enzyme(s).

Sometimes this won't be possible§ — in these cases you could try to find enzymes that leave the same overhang (i.e. have compatible cohesive ends).

There are other techniques that can be done if this isn't possible such as partially filling in the ends to create compatible ends or "blunting" where you fill in and/or chew of the overhangs and then do a blunt end ligation.

Another very common alternative is to use primers with restriction sites at their 5' ends and then PCR amplify the insert you want — this creates a copy of the insert DNA with whatever restriction sites you want at the ends.

§For example there might not be restriction sites for the same enzymes in the correct places in both the vector and insert.

Question 6

What happens to the restriction enzyme once the recombinant plasmid has been formed. Is it destroyed?

Accepted Answer

You must remove or destroy the restriction enzymes (REs) before you ligate.  Otherwise the REs will just recut your newly ligated DNA.

This is often done by purifying the cut DNA — usually by running the digest (cut DNA) on an agarose gel and then cutting out the band of interest.

Question 7

I am quite confused as to the strand of the target gene, when we are cutting the gene are we pasting the whole gene into the plasmid or are we just pasting the part that the restriction enzyme has cut from the whole targeted gene?

Accepted Answer

We are not exactly "pasting" the whole gene, by which I mean that we are not applying ligase to the entire length of the gene. We are _inserting_ the gene, and to do that we are using ligase to paste the two ends cut by the restriction enzymes to the two ends of the plasmid.

Question 8

What if there are not restriction enzymes on either side of the target DNA? Would it still be possible to use restriction enzymes?

Accepted Answer

First, most vectors will have a region known as the "Multiple Cloning Site" (MCS) that can be cut with many different restriction enzymes† — this gives you more choices of enzyme and makes it more likely that you can find one that cuts near the ends of the region you wish to clone.

Second, we often don't care if we clone a small amount of extra DNA , this means that we can search over a larger area than you might expect to find appropriate restriction enzymes.

If the regions flanking the sequence you want to clone don't contain any useful restriction sites you can use primers with restriction sites added to their 5' ends and then amplify the sequence using PCR§.

This amplifies the insert you want and creates a copy of the insert DNA with whatever restriction sites you want added at the ends.

There are many more tricks that have been developed, but adding sites at the ends of primers almost always works, so that is a very good one to know!

Does that help?

†Note: There are hundreds of commercially available restriction enzymes recognizing many different sequences (many of which are palindromes, but not all).

Among these the most commonly used are six-cutters (with 6 bp recognition sites — if you make a bunch of simplifying assumptions you can calculate that these enzymes on average will cut once every 4096 bp.

§Note: Polymerase chain reaction — you can learn more about this technique here:
https://www.khanacademy.org/science/biology/biotech-dna-technology#dna-sequencing-pcr-electrophoresis

Question 9

If you put a same restriction enzymes to two samples of the same person's DNA. The resulting DNA strands after the restriction enzymes cutting should be the same size, right?

So, Can a forensic scientist use gel electrophoresis after this to determine if the DNA of the suspect matches the DNA found or to determine if the found DNA belongs to the criminal or the victim?

Accepted Answer

That is true, but for a typical restriction digest of human DNA you will get around a million different bands with a range of different sizes§ — on a gel this just looks like a smear of DNA and is of no use in identifying individuals.

In addition, two DNA molecules could be exactly the same size, but have different sequences — even if you isolated them (or detected them with a specific probe) you would probably not be able to distinguish them using any form of gel electrophoresis.

Individual humans are about 99.9% identical at the nucleotide level, so telling us apart by DNA requires relatively sophisticated techniques!

There are many different techniques to get information from this sea of DNA, if you want to know more about how this is currently done:
https://en.wikipedia.org/wiki/DNA_profiling

§ ~6.4 billion base pairs in a diploid human genome and a typically six cutter enzyme will on (a very rough) average cut every 4096 bp.

Question 10

Dystrophin is one of the longest genes, with 2.4 million base pairs. Does size have any impact on the size of the plasmid that needs to be used (does it have to be big enough to be able to cut a 2.4 million base pair section out of it?), does the plasmid simply expand to accomodate the gene? Isn't it quite likely that the gene itself would be cut up by the restriction enzymes? How could you assure that the gene would remain in tact and recircularize in the plasmid successfully with such a large gene?

Accepted Answer

A typical plasmid can accommodate inserts of any size up to total size of around 50 kb, but plasmids that are more than 20 kb are very difficult to work with and may require special transformation techniques.  The efficiency of ligation and transformation tends to decrease with extremely large inserts.

For large inserts there are different kinds of vectors (not plasmids) that can be used.

For an enormous insert like you are asking about you would need to use a type of vector known as an artificial chromosome.  These are specific to the type of organism in which you wish to grow the vector with insert.  For example you could use a YAC (yeast artificial chromosome) for the dystrophin gene.

To clone the entire dystrophin gene you would probably have to screen through a library of large fragments made by cloning randomly made fragments of genomic DNA (for example this can be done mechanically by passing the DNA solution through a needle).

In practice you would probably get clones from an already made cDNA library (made by reverse transcribing mRNA) — this means you wouldn't need such a giant vector.  There are actually many different transcripts made from the dystrophin gene that produce different versions of the protein (known as isoforms) — the transcripts range from about 5-14 kb, much more manageable!

If you want to get an idea of the complexity of transcription from the dystrophin gene:
http://www.dmd.nl/isoforms.html

Name	Source	Recognition site	Cut pattern
EcoRI	Escherichia coli	$\begin{aligned} 5^{'} - GAATTC - 3^{'} \\ 3^{'} - CTTAAG - 3^{'} \end{aligned}$ ‍	$\begin{aligned} 5^{'} - G ↓ AATTC - 3^{'} \\ 3^{'} - CTTAA ↑ G - 5^{'} \end{aligned}$ ‍
BamHI	Bacillus amyloliquefaciens	$\begin{aligned} 5^{'} - GGATCC - 3^{'} \\ 3^{'} - CCTAGG - 3^{'} \end{aligned}$ ‍	$\begin{aligned} 5^{'} - G ↓ GATCC - 3^{'} \\ 3^{'} - CCTAG ↑ G - 5^{'} \end{aligned}$ ‍
HindIII	Haemophilus influenzae	$\begin{aligned} 5^{'} - AAGCTT - 3 \\ 3^{'} - TTCGAA - 5^{'} \end{aligned}$ ‍	$\begin{aligned} 5^{'} - A ↓ AGCTT - 3 \\ 3^{'} - TTCGA ↑ A - 5^{'} \end{aligned}$ ‍
SmaI	Serratia marcescens	$\begin{aligned} 5^{'} - CCCGGG - 3 \\ 3^{'} - GGGCCC - 5^{'} \end{aligned}$ ‍	$\begin{aligned} 5^{'} - CCC ↓ GGG - 3 \\ 3^{'} - GGG ↑ CCC - 5^{'} \end{aligned}$ ‍
SacI	Streptomyces achromogenes	$\begin{aligned} 5^{'} - GAGCTC - 3 \\ 3^{'} - CTCGAG - 5^{'} \end{aligned}$ ‍	$\begin{aligned} 5^{'} - GAGCT ↓ C - 3 \\ 3^{'} - C ↑ TCGAG - 5^{'} \end{aligned}$ ‍

Course: Biology library > Unit 20

Restriction enzymes & DNA ligase

Key points:

How do you cut and paste DNA?

Restriction enzymes

DNA ligase

Example: Building a recombinant plasmid

Restriction digests and ligations involve many molecules of DNA

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